When i tried to find alternatives to the shame of the giant Bristol sudoku, it turned out that there are 23 unique sudokus, that you can get by changing 4 digits (which keeps the nice pattern). According to susser this is the hardest one (a swordfish, 2 6-link and a 7-link forcing chain). Are there more elegant solutions ?

Thanks Carcul, does it solve the puzzle for you ?Bob Hansons solver showed me the marked cells, which allow the elimination of 17 in r4c2 (what solves the puzzle). But i dont know how to argue it as an ALS.

On the other hand, as there are many simpler moves that don't achive solution, I was wondering if that solver actually looks for "The Best next move", skipping those non productive essential simpler moves

Wolfgang wrote:Bob Hansons solver showed me the marked cells, which allow the elimination of 17 in r4c2 (what solves the puzzle). But i dont know how to argue it as an ALS..............A and B have a common cell (r5c2):A=1257 in r5c123B=12579 in r2356 (sic)

Sets A and B cannot have a common cell, and Bob Hanson's posts (and I suspect solver too) rarely has an error ... so I think something got lost in transcription. But here's an illustration for the ALS Havard identified above.

tarek wrote:On the other hand, as there are many simpler moves that don't achive solution, I was wondering if that solver actually looks for "The Best next move", skipping those non productive essential simpler moves

No, it did not, it was me, who skipped the other moves. Also it came 2 times with the same sets for eliminating 1 and 7 and had an extra move between them.

But i still think, my reasoning above is correct, isnt it ? So is it no ALS because of the common cell, but the argumentation is the same ?PS: As i see it, it must only hold that x is not in the common cell.

Wolfgang wrote:But i still think, my reasoning above is correct, isnt it ? So is it no ALS because of the common cell, but the argumentation is the same ?............A and B have a common cell (r5c2):A=1257 in r5c123B=12579 in r23561 and 7 in r4c2 both lock 5 in A (r5c1) and in B (r5c2)

For your defined sets, we know either set A or set B cannot contain the 5. When A doesn't contain a 5, A is locked to 127. When B doesn't contain a 5, B is locked to 1279. So any 1s (or alternately, the 2s and 7s) candidates that "see" all 1s (or 2s and 7s) of both sets may be excluded. That sure sounds like the ALS xz-rule to me, where x=5 and z is alternately 1, 2, and 7. And it would hold even if the 5s were weakly linked.

Maybe the sets can share cells as long as the shared cell doesn't contain 'x'. If true, that would certainly open up additional exclusions from the ALS xz-rule.

In any case, your reasoning for this particular case is definitely valid IMO, despite the common cell.

Is this a valid ALS reasoning ?A and B have a common cell (r5c2):A=1257 in r5c123B=12579 in r23561 and 7 in r4c2 both lock 5 in A (r5c1) and in B (r5c2)

The conclusion is certainly correct. Beautiful! This case is more subtle than what we have seen earlier.Do the standard rewrite where no A and B are distinguished. The set is r2356c2,r5c13 of size 6. Digits 12579 with max multiplicities 22121 for a total of 8, so each choice that eliminates three digits can be removed. Now each of the three possibilities for (5,2) already eliminates one digit, and a choice (4,2)1 or (4,2)7 eliminates two further digits, impossible.